IIT Madras
Statistics II
Graded Assignment 4

Graded Assignment 4

Exercise Questions ❓

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Exercise Solutions 🧪

Good morning! Here in India on this Monday, let’s explore these questions. They cover key concepts related to continuous random variables, including Probability Density Functions (PDFs), Cumulative Distribution Functions (CDFs), percentiles, conditional probabilities, and applications of the Normal distribution.

Core Concepts: Continuous Random Variables

  1. Probability Density Function (PDF), $f_X(x)$:

    • Describes the relative likelihood for a continuous random variable to take on a given value.
    • The total area under the PDF curve must equal 1: $\int_{-\infty}^{\infty} f_X(x) dx = 1$.
    • Probabilities are found by integrating the PDF over an interval: $P(a \le X \le b) = \int_{a}^{b} f_X(x) dx$.
    • For any single point $c$, $P(X=c) = 0$.

  2. Cumulative Distribution Function (CDF), $F_X(x)$:

    • Gives the probability that the random variable $X$ is less than or equal to a certain value $x$.
    • $F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) dt$.
    • $P(a < X \le b) = F_X(b) - F_X(a)$.
    • $P(X > a) = 1 - P(X \le a) = 1 - F_X(a)$.
    • CDFs are non-decreasing, right-continuous, $\lim_{x\to-\infty} F_X(x) = 0$, and $\lim_{x\to\infty} F_X(x) = 1$.
    • Jumps in a CDF indicate discrete probability mass at that point.

  3. Percentiles: The $p$-th percentile is the value $k$ such that $P(X \le k) = p/100$, or $F_X(k) = p/100$.

  4. Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

  5. Normal Distribution: A bell-shaped curve defined by its mean ($\mu$) and standard deviation ($\sigma$). We use the standard normal distribution (Z-distribution, $\mu=0, \sigma=1$) and Z-scores ($Z = \frac{X-\mu}{\sigma}$) to calculate probabilities.

Question 1: Minimum Supply Calculation (from file image_7456d4.png)

The Question: A petrol station receives a supply once each day. The daily petrol sales (in thousands of liters) are represented by the random variable $X$, which has the PDF: $f_X(x) = \begin{cases} k(1-x)^4, & 0 \le x \le 1 \ 0, & \text{otherwise} \end{cases}$. What should be the minimum supply (in thousands of liters) so that the probability of the petrol running out on any given day is only 1%?

Detailed Solution:

  1. Find the value of k: The total probability must be 1.

    • $\int_{0}^{1} k(1-x)^4 dx = 1$.
    • Let $u = 1-x$, then $du = -dx$. When $x=0, u=1$. When $x=1, u=0$.
    • $\int_{1}^{0} k u^4 (-du) = k \int_{0}^{1} u^4 du = k \left[\frac{u^5}{5}\right]_0^1 = k(\frac{1}{5} - 0) = \frac{k}{5}$.
    • So, $\frac{k}{5} = 1 \implies k=5$.
    • The PDF is $f_X(x) = 5(1-x)^4$ for $0 \le x \le 1$.

  2. Set up the probability: “Running out” means the demand $X$ is greater than the supply $S$. We want $P(X > S) = 0.01$.

  3. Calculate the probability using the PDF:

    • $P(X > S) = \int_{S}^{1} 5(1-x)^4 dx$.
    • Using the same substitution $u=1-x, du=-dx$. When $x=S, u=1-S$. When $x=1, u=0$.
    • $\int_{1-S}^{0} 5u^4 (-du) = 5 \int_{0}^{1-S} u^4 du = 5 \left[\frac{u^5}{5}\right]_0^{1-S} = [u^5]_0^{1-S} = (1-S)^5 - 0^5 = (1-S)^5$.

  4. Solve for S:

    • $(1-S)^5 = 0.01$
    • $1-S = (0.01)^{1/5}$
    • $1-S \approx 0.3981$
    • $S = 1 - 0.3981 = 0.6019$.

Final Answer: 0.6019 (thousand liters)

Question 2: Probability from CDF (from file image_7456d4.png)

The Question: Let $X$ be a continuous random variable with the CDF: $F_X(x) = \begin{cases} 0, & x < 1 \ \frac{(x-1)^4}{8}, & 1 \le x < 3 \ 1, & x \ge 3 \end{cases}$. What is the value of $P(1 < X < 2)$?

Detailed Solution:

  1. Apply the CDF property: For a continuous variable, $P(a < X < b) = P(a < X \le b) = F_X(b) - F_X(a)$.
  2. Use the given values: $a=1, b=2$.
    • $P(1 < X < 2) = F_X(2) - F_X(1)$.
  3. Evaluate the CDF at the limits:
    • For $F_X(2)$, we use the middle rule since $1 \le 2 < 3$: $F_X(2) = \frac{(2-1)^4}{8} = \frac{1^4}{8} = \frac{1}{8}$.
    • For $F_X(1)$, we use the middle rule since $1 \le 1 < 3$: $F_X(1) = \frac{(1-1)^4}{8} = \frac{0^4}{8} = 0$.
  4. Calculate the probability:
    • $P(1 < X < 2) = \frac{1}{8} - 0 = \frac{1}{8}$.
  5. Convert to decimal: $1 \div 8 = 0.125$.

Final Answer: 0.125

Questions 3, 4, & 5: Probabilities from a Step CDF (from file image_745789.png)

Common Data: Let the CDF of a random variable $X$ be given by: $F_X(x) = \begin{cases} 0, & \text{for } x < 0 \ 1/3, & \text{for } 0 \le x < 1 \ 1/2, & \text{for } 1 \le x < 2 \ 1, & \text{for } x \ge 2 \end{cases}$.

Core Concept: Jumps in a CDF represent discrete probability masses. The probability at a specific point $c$ is the size of the jump: $P(X=c) = F_X(c) - \lim_{x \to c^-} F_X(x)$.

3) Find $P(X=0)$.

Detailed Solution:

  • $P(X=0) = F_X(0) - \lim_{x \to 0^-} F_X(x)$
  • $F_X(0)$ uses the rule $0 \le x < 1$, so $F_X(0) = 1/3$.
  • $\lim_{x \to 0^-} F_X(x)$ uses the rule $x < 0$, so the limit is 0.
  • $P(X=0) = 1/3 - 0 = 1/3$.
  • $1/3 \approx 0.333…$

Final Answer: 0.33

4) Find $P(X=1)$.

Detailed Solution:

  • $P(X=1) = F_X(1) - \lim_{x \to 1^-} F_X(x)$
  • $F_X(1)$ uses the rule $1 \le x < 2$, so $F_X(1) = 1/2$.
  • $\lim_{x \to 1^-} F_X(x)$ uses the rule $0 \le x < 1$, so the limit is 1/3.
  • $P(X=1) = 1/2 - 1/3 = 3/6 - 2/6 = 1/6$.
  • $1/6 \approx 0.166…$

Final Answer: 0.17

5) Find $P(X=2)$.

Detailed Solution:

  • $P(X=2) = F_X(2) - \lim_{x \to 2^-} F_X(x)$
  • $F_X(2)$ uses the rule $x \ge 2$, so $F_X(2) = 1$.
  • $\lim_{x \to 2^-} F_X(x)$ uses the rule $1 \le x < 2$, so the limit is 1/2.
  • $P(X=2) = 1 - 1/2 = 1/2$.
  • $1/2 = 0.5$.

Final Answer: 0.5

Question 6: Percentiles (from file image_7453eb.png)

The Question: The PDF of a continuous random variable X is given by: $f_X(x) = \begin{cases} kx, & 0 \le x < 2 \ 0, & \text{otherwise} \end{cases}$. Find the 75th percentile of the random variable X.

Detailed Solution:

  1. Find k: The total area under the PDF must be 1.

    • $\int_{0}^{2} kx dx = 1$
    • $k \left[\frac{x^2}{2}\right]_0^2 = 1 \implies k (\frac{2^2}{2} - 0) = 1 \implies k(2) = 1 \implies k = 1/2$.
    • The PDF is $f_X(x) = \frac{1}{2}x$ for $0 \le x < 2$.

  2. Set up the percentile equation: We need to find the value $v$ such that $P(X \le v) = 0.75$. This is the same as finding $v$ where $F_X(v) = 0.75$.

  3. Calculate the integral:

    • $P(X \le v) = \int_{0}^{v} f_X(x) dx = \int_{0}^{v} \frac{1}{2}x dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_0^v = \frac{1}{4}v^2$.

  4. Solve for v:

    • $\frac{1}{4}v^2 = 0.75 = \frac{3}{4}$
    • $v^2 = 3$
    • Since the domain is $0 \le x < 2$, we take the positive root: $v = \sqrt{3}$.

  5. Calculate the decimal value: $v = \sqrt{3} \approx 1.73205$.

Final Answer: 1.732

Question 7: Conditional Probability (from file image_7453eb.png)

The Question: Let X be a continuous random variable with PDF: $f_X(x) = \begin{cases} 5x^4 & 0 < x \le 1 \ 0 & \text{otherwise} \end{cases}$. Find $P(X < 8/14 \mid X > 1/4)$.

Detailed Solution:

  1. Apply the conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
    • Event A: $X < 8/14 = 4/7$.
    • Event B: $X > 1/4$.
    • Event $A \cap B$: $1/4 < X < 4/7$.
  2. Calculate $P(B) = P(X > 1/4)$:
    • $P(X > 1/4) = \int_{1/4}^{1} 5x^4 dx = 5 \left[\frac{x^5}{5}\right]{1/4}^{1} = [x^5]{1/4}^{1} = 1^5 - (1/4)^5 = 1 - \frac{1}{1024} = \frac{1023}{1024}$.
  3. Calculate $P(A \cap B) = P(1/4 < X < 4/7)$:
    • $P(1/4 < X < 4/7) = \int_{1/4}^{4/7} 5x^4 dx = [x^5]_{1/4}^{4/7} = (4/7)^5 - (1/4)^5$.
    • $(4/7)^5 = \frac{1024}{16807}$. $(1/4)^5 = \frac{1}{1024}$.
    • $P(A \cap B) = \frac{1024}{16807} - \frac{1}{1024}$.
  4. Calculate $P(A|B)$:
    • $P(A|B) = \frac{\frac{1024}{16807} - \frac{1}{1024}}{\frac{1023}{1024}}$.
    • This requires a calculator:
      • $1024/16807 \approx 0.060927$.
      • $1/1024 \approx 0.00097656$.
      • $1023/1024 \approx 0.9990234$.
      • $P(A \cap B) \approx 0.060927 - 0.00097656 = 0.05995$.
      • $P(A|B) \approx \frac{0.05995}{0.9990234} \approx 0.06001$.

Final Answer: 0.060

Question 8: Normal Distribution Warranty (from file image_7453eb.png)

The Question: A firm produces machines with a lifespan (mean=151 months, std dev=60 months). Find the warranty period to offer so that they replace no more than 15.8% of machines. Use $P(Z < -1) = 0.158$.

Core Concept: We need to find the value $X$ (warranty period) such that the probability of a machine failing before that time, $P(\text{Lifespan} < X)$, is 0.158. We use the Z-score transformation.

Detailed Solution:

  1. Identify the parameters: $\mu = 151$, $\sigma = 60$.
  2. Define the probability: We want $P(\text{Lifespan} < X) = 0.158$.
  3. Convert to Z-score: The Z-score corresponding to $X$ is $Z = \frac{X - \mu}{\sigma} = \frac{X - 151}{60}$.
  4. Rewrite the probability in terms of Z: $P\left(Z < \frac{X - 151}{60}\right) = 0.158$.
  5. Use the given hint: We are told $P(Z < -1) = 0.158$.
  6. Equate the Z-scores: Since the probabilities are equal, the Z-scores must be equal.
    • $\frac{X - 151}{60} = -1$
  7. Solve for X:
    • $X - 151 = -60$
    • $X = 151 - 60 = 91$.

Final Answer: The guarantee period should be 91 months.

Question 9: Finding k for Exponential PDF (from file image_745388.png)

The Question: Let X be a continuous random variable with the PDF: $f_X(x) = \begin{cases} ke^{-x} & x \ge 0 \ 0 & \text{otherwise} \end{cases}$. Find the value of k.

Core Concept: The total area under any PDF must equal 1. $\int_{-\infty}^{\infty} f_X(x) dx = 1$.

Detailed Solution:

  1. Set up the integral:
    • $\int_{0}^{\infty} ke^{-x} dx = 1$.
  2. Evaluate the improper integral:
    • $k \int_{0}^{\infty} e^{-x} dx = k \lim_{b\to\infty} \int_{0}^{b} e^{-x} dx$.
    • $k \lim_{b\to\infty} [-e^{-x}]0^b = k \lim{b\to\infty} (-e^{-b} - (-e^0)) = k \lim_{b\to\infty} (-e^{-b} + 1)$.
    • As $b \to \infty$, $e^{-b} \to 0$.
    • So the integral evaluates to $k(0 + 1) = k$.
  3. Solve for k:
    • Since the integral must equal 1, we have $k=1$.

Final Answer: $k = 1$.

Question 10: Conditional Probability (from file image_745388.png)

The Question: Let X be a continuous random variable with PDF: $f_X(x) = \begin{cases} 5x^4 & 0 < x \le 1 \ 0 & \text{otherwise} \end{cases}$. Find $P(X \le 6/7 \mid X > 1/7)$.

Detailed Solution:

  1. Apply the conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
    • Event A: $X \le 6/7$.
    • Event B: $X > 1/7$.
    • Event $A \cap B$: $1/7 < X \le 6/7$.
  2. Calculate $P(B) = P(X > 1/7)$:
    • $P(X > 1/7) = \int_{1/7}^{1} 5x^4 dx = 5 \left[\frac{x^5}{5}\right]{1/7}^{1} = [x^5]{1/7}^{1} = 1^5 - (1/7)^5 = 1 - \frac{1}{16807} = \frac{16806}{16807}$.
  3. Calculate $P(A \cap B) = P(1/7 < X \le 6/7)$:
    • $P(1/7 < X \le 6/7) = \int_{1/7}^{6/7} 5x^4 dx = [x^5]_{1/7}^{6/7} = (6/7)^5 - (1/7)^5$.
    • $(6/7)^5 = \frac{7776}{16807}$. $(1/7)^5 = \frac{1}{16807}$.
    • $P(A \cap B) = \frac{7776 - 1}{16807} = \frac{7775}{16807}$.
  4. Calculate $P(A|B)$:
    • $P(A|B) = \frac{7775/16807}{16806/16807} = \frac{7775}{16806}$.
  5. Convert to decimal: $7775 \div 16806 \approx 0.46263$.

Final Answer: 0.463

Questions 11-16: Fill-in-the-Blanks with CDF (from files image_74532a.png, image_745029.png)

The Question: Let X be a continuous random variable with the CDF… If $P(X \le 4) = 3/4$, then find the value of $P(1 \le X \le 5)$. Fill in the blanks A-F.

Detailed Solution:

  • Find k (A, B): We are given $P(X \le 4) = F(4) = 3/4$.

    • Since $2 \le 4 \le 6$, we use the middle part of the CDF: $F(4) = k + \frac{4-2}{8} = k + \frac{2}{8} = k + \frac{1}{4}$.
    • Set this equal to the given probability: $k + \frac{1}{4} = \frac{3}{4}$.
    • Solve for k: $k = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.
    • Blank A represents the equation $k+\frac{1}{4} = \frac{3}{4}$. The value $\frac{3}{4}$ is option (11).
    • Blank B represents the value of $k$. $k = \frac{1}{2}$, which is option (10).

  • Calculate $P(1 \le X \le 5)$ (C, D, E, F):

    • For continuous variables, $P(a \le X \le b) = P(X \le b) - P(X < a)$. Since $P(X=a)=0$, this is the same as $P(X \le b) - P(X \le a) = F(b) - F(a)$.
    • So, $P(1 \le X \le 5) = P(X \le 5) - P(X \le 1) = F(5) - F(1)$.
    • Blank C represents $P(X \le 5)$, which corresponds to option (1).
    • Blank D represents $P(X \le 1)$, which corresponds to option (2).
    • Now evaluate $F(5)$ and $F(1)$ using $k=1/2$:
      • Since $2 \le 5 \le 6$, $F(5) = k + \frac{5-2}{8} = \frac{1}{2} + \frac{3}{8} = \frac{4}{8} + \frac{3}{8} = \frac{7}{8}$.
      • Since $0 \le 1 < 2$, $F(1) = \frac{k \cdot 1^2}{4} = \frac{(1/2)}{4} = \frac{1}{8}$.
    • Blank E represents the value of $F(5)$, which is $\frac{7}{8}$, option (8).
    • Calculate the final probability: $P(1 \le X \le 5) = \frac{7}{8} - \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$.
    • Blank F represents the final answer, which is $\frac{3}{4}$, option (11).

Final Answers:

  • 11) A: 11
  • 12) B: 10
  • 13) C: 1
  • 14) D: 2
  • 15) E: 8
  • 16) F: 11
Graded Assignment 3Graded Assignment 5